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I am trying to calculate the energy it would take to destroy a small celestial object made solely of silica. A lot of literature have defined the energy required to shatter a celestial object as the energy required such that the largest aggregate is half that of the original object.

I have tried going about it from a thermal physics approach and I've calculated the energy required to break all the bonds by considering the silica object to be like a crystal of uniform density of silicon dioxide (which are some major assumptions) as

$E = mol * E_{BDE} * N$

where $mol$ is the moles, $E_{BDE}$ is the bond disassociation energy of one bond of silica and N is the number of bonds in one molecule of silica ($N=2$).

I assume a percentage of that (i.e. $F*E$ where $F$ is a dimensionless percentage) is required to produce a half-sized fragment of the silica object but I am unsure as to how to calculate that. Does anybody have any ideas how a reasonable estimate for $F$ could be obtained?

I've assumed that the object is pretty small (radius of 1000 metres) so that its in the region dominated by material strength rather than graviational strength (ie on the LHS of the figure below).

enter image description here I tried going about it by assuming the object is being split in half (to produce half sized fragments) so that the number of bonds being broken would be proportional to the surface area of the cleaved surface (cross-sectional area, C-SA) and the total number of bonds would be proportional to the volume (V) and hence the percentage would be like C-SA/V but this would not give a dimensionless percentage prefactor.

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  • $\begingroup$ Planets are held together by gravity, not by being one gigantic crystal with atomic bonds. Still might be instructive to calculate this, but I assure you if you do the calculation correctly the gravitational potential energy you need to overcome will be a lot more than the bond breaking. Particularly because the gravitational potential energy is proportional to volume, but the bond breaking energy is proportional to the cross sectional area $\endgroup$
    – AXensen
    Commented Apr 18, 2023 at 15:35
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    $\begingroup$ @AXensen Apologies, I probably wasn't clear in my original question. I've redrafted it now, the size mangitudes I was talking about was on the order of 1000 metres for the radius so more like a comet or asteroid than a planet so that it's energy is in the material strength regime and is dominated by the bond energies. $\endgroup$
    – John
    Commented Apr 18, 2023 at 15:45
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    $\begingroup$ At smaller length scales, so-called “rubble pile” asteroids are more common than giant crystals. $\endgroup$
    – rob
    Commented Apr 18, 2023 at 16:05

2 Answers 2

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The volume of the crystal in which the bonds are broken would be something like the cross-sectional area times the interatomic spacing in the lattice. This would have units of volume, as expected, and you could use it to estimate the "cleavage energy" required via the method you outline above.

This would probably only be correct to within an order of magnitude, but at the very least it would let you compare this number to the gravitational binding energy of the sphere, which (as pointed out in the comments) would probably be significantly larger. Note that for a crystal of fixed density, the gravitational binding energy is proportional to the fifth power of the radius ($U \propto M^2/R \propto \rho R^5$) while the cleavage energy, as described, would only be proportional to $R^2$ (i.e., proportional to the cross-sectional area.) So for a sufficiently large crystal, the gravitational binding energy will win out.

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  • $\begingroup$ @AXensen: I was developing the OP's method: if we can find the fraction $F$ of the bonds that were disrupted, and we know the total bonding energy of the entire crystal (which the OP seems to have a handle on how to calculate), then we can find the energy required to break all of the bonds. And the bonds that are disrupted when we cleave the plane will be those in a thin slice that's about one atomic separation thick and with an area of the cleavage plane itself. (All the bonds outside this volume would remain unaffected by the split.) ... $\endgroup$ Commented Apr 18, 2023 at 19:08
  • $\begingroup$ ... So the fraction of the disrupted bonds would be something like $R^2 a/R^3 \sim a/R$. $\endgroup$ Commented Apr 18, 2023 at 19:12
  • $\begingroup$ yeah I see. sorry about that. deleted my comment. $\endgroup$
    – AXensen
    Commented Apr 18, 2023 at 20:08
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To split this thing in half, which you've declared as your goal, you could estimate the number of atomic bonds in a cross section, which would be the area of a cross section $\pi R^2$ divided by the inter-atomic spacing squared $\pi R^2/a^2$. I don't think you need to break all of, or even a significant fraction of, the bonds in the entire crystal. That seems like a pretty overkill definition of "destruction." Then you just multiply by the bond strength, or for order of magnitude estimates, you could use the typical scale of chemical bonds, $\sim1\text{ eV}$. So I have $(1\text{ eV})\pi R^2/a^2$.

But note that for big enough objects, gravity, not bonds, is what holds them together. The gravitational potential energy of a uniform density sphere is: $$ -\frac{4GM^2}{5R} $$ Where $M=4\pi R^3\rho/3$. Setting these equal to eachother we find the object size such that gravity starts to be the dominant mechanism holding the thing together, which is given by: $$ \frac{4G(4\pi R^3\rho/3)^2}{5R}=(1\text{ eV})\pi R^2/a^2 $$ $$ \frac{4GR^3(4\pi \rho/3)^2}{5}=(1\text{ eV})\pi /a^2 $$ AFter googling silica lattice spacing and silica density we get $$ R=\left(\frac{5}{4G(4\pi \rho/3)^2}(1\text{ eV})\pi /a^2\right)^{1/3}=11\text{ m} $$ Or if I use the propper binding energy in silica of 2.3eV, this increases by ~30% to 14m

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    $\begingroup$ Note that we would expect the total number of bonds in a sphere of radius $R$ to be something like $\frac43 \pi R^3/a^3$, so the fraction of the total number of bonds that will be broken would be $$F = \frac{ \pi R^2/a^2}{\frac43 \pi R^3/a^3} = \frac{3}{4} \frac{a}{R},$$in line with my estimate above. $\endgroup$ Commented Apr 18, 2023 at 19:14
  • $\begingroup$ @MichaelSeifert Thank you very much, I got the same result after seeing the responses on here and it checks out that it's dimensionless which was my initial issue. $\endgroup$
    – John
    Commented Apr 18, 2023 at 19:50

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